Optimal. Leaf size=48 \[ \frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {\sqrt {1-2 x}}{5 (5 x+3)} \]
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Rubi [A] time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {47, 63, 206} \[ \frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {\sqrt {1-2 x}}{5 (5 x+3)} \]
Antiderivative was successfully verified.
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Rule 47
Rule 63
Rule 206
Rubi steps
\begin {align*} \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x}}{5 (3+5 x)}-\frac {1}{5} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {\sqrt {1-2 x}}{5 (3+5 x)}+\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {\sqrt {1-2 x}}{5 (3+5 x)}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 67, normalized size = 1.40 \[ \frac {\sqrt {1-2 x} \left (-110 x+2 \sqrt {55} \sqrt {2 x-1} (5 x+3) \tan ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )+55\right )}{275 \left (10 x^2+x-3\right )} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.02, size = 54, normalized size = 1.12 \[ \frac {\sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, \sqrt {-2 \, x + 1}}{275 \, {\left (5 \, x + 3\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.23, size = 56, normalized size = 1.17 \[ -\frac {1}{275} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {\sqrt {-2 \, x + 1}}{5 \, {\left (5 \, x + 3\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 36, normalized size = 0.75 \[ \frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{275}+\frac {2 \sqrt {-2 x +1}}{25 \left (-2 x -\frac {6}{5}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.20, size = 53, normalized size = 1.10 \[ -\frac {1}{275} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {\sqrt {-2 \, x + 1}}{5 \, {\left (5 \, x + 3\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.17, size = 35, normalized size = 0.73 \[ \frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{275}-\frac {2\,\sqrt {1-2\,x}}{25\,\left (2\,x+\frac {6}{5}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 1.61, size = 175, normalized size = 3.65 \[ \begin {cases} \frac {2 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{275} + \frac {\sqrt {2}}{25 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {11 \sqrt {2}}{250 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} & \text {for}\: \frac {11}{10 \left |{x + \frac {3}{5}}\right |} > 1 \\- \frac {2 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{275} - \frac {\sqrt {2} i}{25 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {11 \sqrt {2} i}{250 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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